I'm interested in the solution to the following integral, given that $a$ is a positive real: $$I=\int_0^1 \text{erfc}\!\left(\frac{a}{\sqrt{t}}\right) \text{erfc}\!\left(\frac{a}{\sqrt{1-t}}\right) dt=(2 b+1) \text{erfc}\left(\sqrt{b}\right)-\frac{2 \sqrt{b} e^{-b}}{\sqrt{\pi }}$$ where $a=\sqrt{\frac{b}{4}}$. The result (that Mariusz pointed out in the comments and i reproduced using mathematica) does not suffice, I require the step by step solution or a reference.
Actually, it seems that this is the final step in a lengthy calculation of the integral in this question: Evaluation of $\int_0^T \int_0^t \tau^{-1/2} (T-\tau)^{-1/2} \exp(\frac{ix^2}{2(T-\tau)})d\tau dt$
Using the convolution theorem for laplace transformations, one gets: $$I=\mathcal{L}^{-1}\left[s^{-2} e^{-2\sqrt{bs}}\right](1)$$ Sadly, i could not find a way to perform this inverse laplace transformation, so I tried to tackle the integral directly. Using $erfc=1-erf$ and dropping some easy to calculate terms, I needed to calculate: $$J=\int_0^1 \text{erf}\!\left(\frac{a}{\sqrt{t}}\right) \text{erf}\!\left(\frac{a}{\sqrt{1-t}}\right) dt=\frac{4}{\pi}\int_0^1\int_0^{\frac{a}{\sqrt{t}}}\int_0^{\frac{a}{\sqrt{1-t}}}e^{-x^2-y^2}dxdydt$$
Or, written a bit differently: $$J=\frac{4}{\pi}\int_0^1\int_{\eta(t)}e^{-v^2}d^2vdt$$ where $\eta(t)=[0,\frac{a}{\sqrt{t}}]\times[0,\frac{a}{\sqrt{1-t}}]$
The function to be integrated does, in this form, exhibit cylindrical symmetry, thus i tried switching into cylindrical coordinates:
$$J=\frac{4}{\pi}\int_0^{+\infty}\int_0^1l(r|\eta(t))dt \exp(-r^2)dr$$ where $l(r|\eta(t))$ is the arc length of the section of a circle of radius $r$ with $\eta(t)$. Some geometric reasoning gives($A\leq B$): $$l(r|[0,A]\times [0,B])=\begin{cases} \frac{\pi}{2}r & r\leq A\\ \frac{\pi}{2}r - r\arctan(\sqrt{\frac{r^2}{A^2}-1}) & A\leq r\leq B\\ \frac{\pi}{2}r - r\arctan(\sqrt{\frac{r^2}{A^2}-1}) -r\arctan(\sqrt{\frac{r^2}{B^2}-1}) & B\leq r\leq \sqrt{A^2+B^2}\\ 0 & \sqrt{A^2+B^2}\leq r \end{cases}$$
Also, it is noteworthy to see: $$l(r|[0,A]\times [0,B])=Re(\frac{\pi}{2}r - r\arctan(\sqrt{\frac{r^2}{A^2}-1}) -r\arctan(\sqrt{\frac{r^2}{B^2}-1})) \Theta(\sqrt{A^2+B^2}-r)$$
The heaviside step function can be taken into account by adjusting the integration range in $t$ depending on $r$: for $r\leq \sqrt{A^2+B^2}$, one integrates from 0 to 1. for $\sqrt{A^2+B^2}\leq r$, one integrates over: $$[0,1/2-\sqrt{1/4-a^2/r^2}]\cup [1/2+\sqrt{1/4-a^2/r^2},1] (*)$$
Luckily, $\arctan\sqrt{\frac{r^2t}{a^2}-1}$ (appearing in $l(r|\eta(t))$) does posses an easy antiderivative(which i will not write out here)! Also, the antiderivative's value at the intermediate points in (*) are purely imaginary, so that taking the real part cancels them.
So finally, if I'm not totally mistaken, we arrive at:
$$J=\frac{4}{\pi}Re\left[\int_0^{+\infty}e^{-r^2}\left( \frac{\pi}{2}r\left(1-2\sqrt{1/4-a^2/r^2}\right) -2a^2/r\left(r^2/a^2\arctan\left(\sqrt{\frac{r^2t}{a^2}-1}\right) -\sqrt{\frac{r^2t}{a^2}-1}\right) \right)dr\right] $$
I think this transformation had no merit for this integral seem even scarier than the one i started with, but maybe someone has an idea how to work with that.
